Use the Mittag-Leffler theorem to prove the following: Let $(a_n)$ be a sequence in a simply connected domain $D \subset \mathbb{C}$ that does not have an accumulation point in $D$. Prove that there exists a holomorphic function $f$ so that its zero set equals $(a_n)$ counting multiplicity.

In particular, do not use the Weierstrass Factorization Theorem, Weierstrass Products, or Blaschke Products in your answer.

I have no idea how to do this problem. Can someone help?

  • $\begingroup$ Do you know what the Mittag-Leffler theorem says? $\endgroup$ – Pedro Tamaroff Mar 31 '16 at 6:10
  • $\begingroup$ The Mittag-Leffler theorem says that it is possible to construct a meromorphic function with the poles and singular parts of your choosing. $\endgroup$ – Nick Mar 31 '16 at 6:12
  • $\begingroup$ that doesn't give the existence of a function with zeros at $(a_n)$ $\endgroup$ – Nick Mar 31 '16 at 6:21

Idea: If $f(a) = 0$ with multiplicity $k$ then the logarithmic derivative $(\log f)' = f'/f$ satisfies $$ \frac{f'(z)}{f(z)} = \frac{k}{z-a} + O(1) \quad \text{ for } z \to a \, . $$

Use the Mittag-Leffler theorem to construct a meromorphic function $g$ in $D$ with poles exactly at the points $a_n$ and principal parts $$ \frac{k_n}{z-a_n} $$ where $k_n$ is the desired multiplicity at $a_n$.

Then show that there is a holomorphic function $f$ in $D$ such that $$ \frac{f'}{f} = g \, ,$$ details can be found for example in Which meromorphic functions are logarithmic derivatives of other meromorphic functions?.

$f$ has zeros exactly at the points $a_n$ with multiplicity $k_n$.

  • $\begingroup$ Doesn't the link you provide only show that $f$ is guaranteed to be meromorphic, not necessarily entire? $\endgroup$ – TSF Apr 9 '16 at 19:41
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    $\begingroup$ @TonyS.F.: Here all $k_n$ are positive , therefore $f$ is entire. If $k_n \in \Bbb Z$ then $f$ can be meromorphic. $\endgroup$ – Martin R Apr 9 '16 at 19:49

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